Let \(X\) and \(Y\) be metric spaces with metrics \(d_X\) and \(d_Y\), respectively. Let \(f: X \to Y\) have the property that for every pair of points \(x_1, x_2\) of \(X\),
\[ d_Y(f(x_1), f(x_2)) = d_X(x_1, x_2). \]
Show that \(f\) is an imbedding. It is called an isometric imbedding of \(X\) in \(Y\).
Solution
If \(f(x_1)=f(x_2)\), then \[ d_X(x_1,x_2)=d_Y(f(x_1),f(x_2))=0, \] so \(x_1=x_2\). Thus \(f\) is injective.
Moreover \[ d_Y(f(x_1),f(x_2))=d_X(x_1,x_2) \] shows directly that \(f\) is continuous: the inverse image of the radius-\(r\) ball in \(f(X)\) about \(f(x)\) is the radius-\(r\) ball in \(X\) about \(x\). The inverse map \[ f^{-1}:f(X)\to X \] has the same property and is therefore continuous. Hence \(f:X\to f(X)\) is a homeomorphism, so \(f\) is an embedding.